What is the integral from 0 to 5 of (18)/(\sqrt[4]{3x+1)} ?

The integral from 0 to 5 of (18)/(\sqrt[4]{3x+1)} is 56

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integral from 0 to 5 of (18)/(\sqrt[4]{3x+1)}

56

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

: [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16}

= 18 \cdot \frac{1}{3} [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16}

Simplify

18 \cdot \frac{1}{3} [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16} : 6 [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16}

= 6 [\frac{4}{3} u^{\frac{3}{4}}]_{1}^{16}

Compute the boundaries:

\frac{28}{3}

= 6 \cdot \frac{28}{3}

Simplify

= 56

\int x cos (5 π) x^{2} d x

\int \frac{9 x - 18}{x ^{3} - 2 x ^{2} - x + 2} d x

\frac{d y}{d x} = \frac{x y + 5 x - y - 5}{x y - 2 x + 6 y - 12}

\frac{d}{d x} ((x^{2} + 5 x + 1) f (x) (sin (3 x)))

\int x^{2} \cdot x - 2 d x

What is the integral from 0 to 5 of (18)/(\sqrt[4]{3x+1)} ?
The integral from 0 to 5 of (18)/(\sqrt[4]{3x+1)} is 56